Answer
When $[S] = 100 mM$, the velocity is close to $V_{max}$. Therefore, assume that $V_{max} \approx 50 μM \cdot s^{−1}$
Using the Michaelis–Menten equation:
$$v_{\mathrm{o}}=\frac{V_{\max }[\mathrm{S}]}{K_{M}+[\mathrm{S}]}$$
$$K_{M}+[\mathrm{S}]=\frac{V_{\max }[\mathrm{S}]}{v_{\mathrm{o}}}$$
$$K_{M}=\frac{V_{\max }[\mathrm{S}]}{v_{\mathrm{o}}}-[\mathrm{S}]$$
$$K_{M}=\frac{\left(50 \mu \mathrm{M} \cdot \mathrm{s}^{-1}\right)(1 \mu \mathrm{M})}{\left(5 \mu \mathrm{M} \cdot \mathrm{s}^{-1}\right)}-(1 \mu \mathrm{M})=9 \mu \mathrm{M}$$
The true $V_{\max }$ must be greater than the estimated value, so the value of $K_{M}$ is an underestimate of the true $K_{M}$ .
Work Step by Step
When $[S] = 100 mM$, the velocity is close to $V_{max}$. Therefore, assume that $V_{max} \approx 50 μM \cdot s^{−1}$
Using the Michaelis–Menten equation:
$$v_{\mathrm{o}}=\frac{V_{\max }[\mathrm{S}]}{K_{M}+[\mathrm{S}]}$$
$$K_{M}+[\mathrm{S}]=\frac{V_{\max }[\mathrm{S}]}{v_{\mathrm{o}}}$$
$$K_{M}=\frac{V_{\max }[\mathrm{S}]}{v_{\mathrm{o}}}-[\mathrm{S}]$$
$$K_{M}=\frac{\left(50 \mu \mathrm{M} \cdot \mathrm{s}^{-1}\right)(1 \mu \mathrm{M})}{\left(5 \mu \mathrm{M} \cdot \mathrm{s}^{-1}\right)}-(1 \mu \mathrm{M})=9 \mu \mathrm{M}$$
The true $V_{\max }$ must be greater than the estimated value, so the value of $K_{M}$ is an underestimate of the true $K_{M}$ .
